The Legendre's differential equation $$(1-x^2)y''-2xy'+n(n+1)y=0$$
Substitute $y=\sum_{m=0}^\infty a_mx^m$ $$ \sum_{m=2}^\infty m(m-1)a_mx^{m-2}-\sum_{m=2} ^\infty m(m-1)a_mx^m-\sum_{m=1}^\infty 2ma_mx^m+\sum_{x=0}^\infty ka_mx^m=0\\ $$ Set $m=s+2$ $$ \sum_{s=0}^\infty (s+2)(s+1)a_{s+2}x^s-\sum_{s=2}^\infty s(s-1)a_sx^s-\sum_{s=1}^\infty 2sa_sx^s+\sum_{s=0}^\infty ka_sx^s=0\\ \implies \boxed{ a_{s+2}=\frac{-(n-s)(n+s+1)}{(s+2)(s+1)}a_s }\\ \implies y(x)=a_0y_1(x)+a_1y_2(x)\\ $$ where $$ y_1(x)=1-\frac{n(n+1)}{2!}x^2+\frac{(n-1)n(n+1)(n+3)}{4!}x^4-\cdots\\ y_2(x)=x-\frac{(n-1)(n+2)}{3!}x^3+\frac{(n-3)(n-1)(n+2)(n+4)}{5!}x^5-\cdots\\ $$ When $n=s\geq 0$, we have $a_{n+2}=a_{n+4}=\cdots=0$
$$ a_{n-2}=\frac{-n(n-1)}{2(2n-1)}a_n\\ a_{n-4}=\frac{-(n-2)(n-3)}{4(2n-3)}a_{n-2}=(-1)^2\frac{n(n-1)(n-2)(n-3)}{2.4(2n-1)(2n-3)}a_n $$ and so on
It is said that the Legendre polynomials are defined by choosing $\color{red}{a_n=\frac{(2n-1)(2n-3)\cdots3.1}{n!}=\frac{(2n)!}{2^n(n!)^2}}$
I think it comes from $P_n(1)=1$ but how do I derive it from the recurrence relation here ?
My Attempt
$$ a_{n-2m}=(-1)^{m}\frac{n(n-1)(n-2)\cdots(n-(2m-1))}{2.4.\cdots 2m.(2n-1)(2n-3)\cdots(2n-(2m-1))}a_n\\ $$
Set $n=2m$ and $P_0(1)=a_0=1$ $$ a_0=(-1)^{m}.\frac{n!}{2^m.m!.(2n-1)(2n-3)\cdots(n+1)}.a_n\\ a_n=(-1)^{m}.\frac{2^m.m!.(2n-1)(2n-3)\cdots(n+1)}{n!}.a_0\\ =(-1)^{m}.\frac{2^m.m!.(2n-1)(2n-3)\cdots(n+1).(n-1)(n-3)\cdots 5.3.1}{n!.(n-1)(n-3)\cdots 5.3.1}.a_0\\ =(-1)^{m}.\color{red}{ \frac{(2n-1)(2n-3)\cdots3.1}{n!} }.\frac{2^m.m!}{(n-1)(n-3)\cdots 5.3.1}.1 $$ I don't think my derivation is leading to the required choice of the coefficient of the highest degreee. How do I approch the problem then ?
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